The reaction that takes place when Cl_2 gas i | vedclass

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(D) When $Cl_2$ gas is passed through a hot and concentrated $NaOH$ solution, the reaction is: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$ In

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Disproportionation

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(D) When $Cl_2$ gas is passed through a hot and concentrated $NaOH$ solution, the reaction is: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$ In this reaction, the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $NaCl$ (reduction) and $+5$ in $NaClO_3$ (oxidation). Since the same element is simultaneously oxidized and reduced, this is a disproportionation reaction.

The reaction that takes place when $Cl_2$ gas is passed through concentrated $NaOH$ solution is:

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